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18 Apr

### A Representation of Rational Numbers

One of the most amusing instances of indirect proof - proof by contradiction - is the establishment of the existence of two irrational numbers $a$ and $b$ such that $a^b$ is rational. Indeed, $\sqrt{2}$ is irrational. Then if $\sqrt{2}^\sqrt{2}$ is rational the problem is solved with $a=b=\sqrt{2}$. Otherwise, it is solved with $a=\sqrt{2}^\sqrt{2}$ and $b=\sqrt{2}$ because then $a^b=2$.

Stan Dolan has recently proved that practically all rational numbers in the interval $(1/e,\infty)$ are in the form $a^a$, where $a$ is irrational.

The claim is really elementary if we consider a smaller interval $(1,\infty)$. On this interval function $f(x)=x^x$ is clearly monotone increasing. To prove that this is also true for $x\gt 1/e$ requires a little calculus: $f'(x) = (1+\mbox{ln}x)x^x$ and is positive for $x\gt 1/e$.

Function $f$ maps continuously interval $(1/e,\infty)$ onto interval $I=({(1/e)}^{1/e},\infty)$. Let $r\in I$ be rational. There is $a$ such that $r=a^a$. Assume $a$ is also rational. More accurately, assume $r =\frac{b}{c}$ and $a=\frac{n}{m}$, both fractions in lowest terms. Then ${(n/m)}^{n/m}=b/c$. This implies

$n^{n}c^{m}=m^{n}b^m$.

If $m=1$, $r=n^n$. We shall show that this is an exceptional case. In other words, an assumption that $m\gt 1$ leads to a contradiction. Since $m$ and $n$ are mutually prime, every factor of $m^n$ divides $c^m$. The converse is also true because $c$ and $b$ are also mutually prime. It follows that

$c^m=m^n$.

Assume $p$ is a common prime factor of $c$ and $m$. We'll write

$c=p^{i}k$ and $m=p^{j}l$

where both $k$ and $l$ are prime to $p$. Substitution gives

$p^{im}k^m=p^{jn}l^n$.

In particular, $im=jn$. Further, $p^{j}|m$ so that $p^{j}|jn$. But $m$ and $n$ are assumed to be mutually prime, implying $p^{j}|j$. We thus led to

$2^{j}\le p^{j}\le j$.

But this is a contradiction, for, for all positive integer $j$, $2^{j}\gt j$.

To sum up, the identity $a^a=r$, with a rational $r$ is possible in two cases only: when $a$ is an integer or irrational. Put differently, every rational number in the interval $I$, except for those in the form $n^n$, with integer $n$, are in the form $a^a$ with $a$ irrational.

### References

#### 3 Responses to “A Representation of Rational Numbers”

1. 1
Matrix67: My Blog » Blog Archive » 经典证明：几乎所有有理数都是无理数的无理数次方 Says:

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经典证明：几乎所有有理数都是无理数的无理数次方 | Mining Time (Digests) Says:

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经典证明：几乎所有有理数都是无理数的无理数次方 | 好豆饼 Says:

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